integration cos^2(∏/2cosθ)

doey

i facing a maths problem in integrating ∫ cos^2(∏/2cosθ) with limit from 0 to ∏/2,i was panic and struggled a long period of time in solving this,anyone can help me? pls give me the answer in detail tq !

Millennial

Use the integral identity [itex]\displaystyle \int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx[/itex].

arildno

Why would that help, Millenial?
You change the internal cos(theta) to a sin(theta)..

micromass

First of all, do you mean

(1) [itex]\int_0^{\pi/2} \cos^2 (\frac{\pi}{2} \cos\theta) d\theta[/itex]

or

(2) [itex]\int_0^{\pi/2} \cos^2 (\frac{\pi}{2\cos \theta}) d\theta[/itex]

Also, which methods do you have at your disposal? Contour integration? Differentiation under the integral sign? Just normal calc II techniques?

doey

(1) [itex]\int_0^{\pi/2} \cos^2 (\frac{\pi}{2} \cos\theta) d\theta[/itex]
,i am asking this pls let me know the steps it takes

micromass

Which course is this for? Do you know Bessel functions?? The solution requires this (at least that is what wolfram alpha says).

arildno

Just a thought:
We may easily rewrite this equation into the identity:
[tex]\int_{0}^{\frac{\pi}{2}}\cos^{2}(\frac{\pi}{2}\cos\theta)d\theta+\int_{ 0}^{\frac{\pi}{2}}\sin^{2}( \frac{\pi}{2}\cos\theta)d\theta=\frac{\pi}{2}[/tex]
I feel dreadfully tempted to declare the two integrals to have the same value (the latter being merely a flipped version of the first), but temptation is not proof..

micromass

Hmmm, looking at the graph doesn't really convince me that the integrals are equal

Anyway, wolfram alpha gives us

[itex]\int_0^{\pi/2} \cos^2(\frac{\pi}{2} \cos(x))dx = \frac{\pi}{4}(1+J_0(\pi))[/itex]

so I doubt the integral will be solvable with methods like these.

Mute

To evaluate the integral one will have to use the identities

$$\cos t = \frac{1}{2}(e^{it}+e^{-it})$$
(or just ##\cos t = \mbox{Re}[\exp(it)]##)

and

$$e^{iz\cos\theta} = \sum_{n=-\infty}^\infty i^n J_n(z)e^{in\theta}.$$

I guess the trig identity

$$\cos^2 t = \frac{1}{2}(1+\cos(2t))$$
also helps.